1. Splash Screen

2. Five-Minute Check (over Lesson 4–5) CCSS Then/Now New Vocabulary
Key Concept: Quadratic Formula
Example 1: Two Rational Roots
Example 2: One Rational Root
Example 3: Irrational Roots
Example 4: Complex Roots
Key Concept: Discriminant
Example 5: Describe Roots
Concept Summary: Solving Quadratic Equations
Lesson Menu

3. Solve x2 – 2x + 1 = 9 by using the Square Root Property.
C. 2, –2
D. 2, 1
5-Minute Check 1

4. Solve x2 – 2x + 1 = 9 by using the Square Root Property.
C. 2, –2
D. 2, 1
5-Minute Check 1

5. Solve 4c2 + 12c + 9 = 7 by using the Square Root Property.D.
5-Minute Check 2

6. Solve 4c2 + 12c + 9 = 7 by using the Square Root Property.D.
5-Minute Check 2

7. Find the value of c that makes the trinomial x2 + x + c a perfect square. Then write the trinomial as a perfect square. A. B. C.
D. 1; (x + 1)2
5-Minute Check 3

8. Find the value of c that makes the trinomial x2 + x + c a perfect square. Then write the trinomial as a perfect square. A. B. C.
D. 1; (x + 1)2
5-Minute Check 3

9. Solve x2 + 2x + 24 = 0 by completing the square.D.
5-Minute Check 4

10. Solve x2 + 2x + 24 = 0 by completing the square.D.
5-Minute Check 4

11. Solve 5g2 – 8 = 6g by completing the square.
D. 4, 2
5-Minute Check 5

12. Solve 5g2 – 8 = 6g by completing the square.
D. 4, 2
5-Minute Check 5

13. Find the value(s) of k in x2 + kx = 0 that would make the left side of the equation a perfect square trinomial.
A. –10, 10
B. 5, 20
C. –20, 20
D. 4, 25
5-Minute Check 6

14. Find the value(s) of k in x2 + kx = 0 that would make the left side of the equation a perfect square trinomial.
A. –10, 10
B. 5, 20
C. –20, 20
D. 4, 25
5-Minute Check 6

15. Mathematical Practices
Content Standards
N.CN.7 Solve quadratic equations with real coefficients that have complex solutions.
A.SSE.1.b Interpret complicated expressions by viewing one or more of their parts as a single entity.
Mathematical Practices
8 Look for and express regularity in repeated reasoning.
CCSS

16. You solved equation by completing the square.
Solve quadratic equations by using the Quadratic Formula.
Use the discriminant to determine the number and type of roots of a quadratic equation.
Then/Now

17. Quadratic Formula
discriminant
Vocabulary

18. Concept

19. Solve x2 – 8x = 33 by using the Quadratic Formula.
Two Rational Roots
Solve x2 – 8x = 33 by using the Quadratic Formula.
First, write the equation in the form ax2 + bx + c = 0 and identify a, b, and c.
ax2 + bx + c = 0
x2 – 8x = 33 1x2 – 8x – 33 = 0
Then, substitute these values into the Quadratic Formula.
Quadratic Formula
Example 1

20. Replace a with 1, b with –8, and c with –33.
Two Rational Roots
Replace a with 1, b with –8, and c with –33.
Simplify.
Simplify.
Example 1

21. or Write as two equations.
Two Rational Roots
or Write as two equations.
x = x = –3 Simplify.
Answer:
Example 1

22. or Write as two equations.
Two Rational Roots
or Write as two equations.
x = x = –3 Simplify.
Answer: The solutions are 11 and –3.
Example 1

23. Solve x2 + 13x = 30 by using the Quadratic Formula.
Example 1

24. Solve x2 + 13x = 30 by using the Quadratic Formula.
Example 1

25. Solve x2 – 34x + 289 = 0 by using the Quadratic Formula.
One Rational Root
Solve x2 – 34x = 0 by using the Quadratic Formula.
Identify a, b, and c. Then, substitute these values into the Quadratic Formula.
Quadratic Formula
Replace a with 1, b with –34, and c with 289.
Simplify.
Example 2

26. One Rational Root
Answer:
Example 2

27. Answer: The solution is 17.
One Rational Root
Answer: The solution is 17.
Check A graph of the related function shows that there is one solution at x = 17.
[–5, 25] scl: 1 by [–5, 15] scl: 1
Example 2

28. Solve x2 – 22x + 121 = 0 by using the Quadratic Formula.
Example 2

29. Solve x2 – 22x + 121 = 0 by using the Quadratic Formula.
Example 2

30. Solve x2 – 6x + 2 = 0 by using the Quadratic Formula.
Irrational Roots
Solve x2 – 6x + 2 = 0 by using the Quadratic Formula.
Quadratic Formula
Replace a with 1, b with –6, and c with 2.
Simplify. or
Example 3

31. Irrational Roots
Answer:
Example 3

32. Irrational Roots
Answer:
Check Check these results by graphing the related quadratic function, y = x2 – 6x + 2. Using the ZERO function of a graphing calculator, the approximate zeros of the related function are 0.4 and 5.6.
[–10, 10] scl: 1 by [–10, 10] scl: 1
Example 3

33. Solve x2 – 5x + 3 = 0 by using the Quadratic Formula.
Example 3

34. Solve x2 – 5x + 3 = 0 by using the Quadratic Formula.
Example 3

35. Solve x2 + 13 = 6x by using the Quadratic Formula.
Complex Roots
Solve x = 6x by using the Quadratic Formula.
Quadratic Formula
Replace a with 1, b with –6, and c with 13.
Simplify.
Simplify.
Example 4

36. Complex Roots
Answer:
Example 4

37. Answer: The solutions are the complex numbers 3 + 2i and 3 – 2i.
Complex Roots
Answer: The solutions are the complex numbers 3 + 2i and 3 – 2i.
A graph of the related function shows that the solutions are complex, but it cannot help you find them.
[–5, 15] scl: 1 by [–5, 15] scl: 1
Example 4

38. x2 + 13 = 6x Original equation
Complex Roots
Check To check complex solutions, you must substitute them into the original equation. The check for 3 + 2i is shown below.
x = 6x Original equation
(3 + 2i) = 6(3 + 2i) x = (3 + 2i) ?
9 + 12i + 4i = i Square of a sum; Distributive Property ?
i – 4 = i Simplify. ?
i = i 
Example 4

39. Solve x2 + 5 = 4x by using the Quadratic Formula.
A. 2 ± i
B. –2 ± i
C i
D. –2 ± 2i
Example 4

40. Solve x2 + 5 = 4x by using the Quadratic Formula.
A. 2 ± i
B. –2 ± i
C i
D. –2 ± 2i
Example 4

41. Concept

42. b2 – 4ac = (3)2 – 4(1)(5) Substitution = 9 – 20 Simplify.
Describe Roots
A. Find the value of the discriminant for x2 + 3x + 5 = 0. Then describe the number and type of roots for the equation.
a = 1, b = 3, c = 5
b2 – 4ac = (3)2 – 4(1)(5) Substitution
= 9 – 20 Simplify.
= –11 Subtract.
Answer:
Example 5

43. b2 – 4ac = (3)2 – 4(1)(5) Substitution = 9 – 20 Simplify.
Describe Roots
A. Find the value of the discriminant for x2 + 3x + 5 = 0. Then describe the number and type of roots for the equation.
a = 1, b = 3, c = 5
b2 – 4ac = (3)2 – 4(1)(5) Substitution
= 9 – 20 Simplify.
= –11 Subtract.
Answer: The discriminant is negative, so there are two complex roots.
Example 5

44. b2 – 4ac = (–11)2 – 4(1)(10) Substitution = 121 – 40 Simplify.
Describe Roots
B. Find the value of the discriminant for x2 – 11x + 10 = 0. Then describe the number and type of roots for the equation.
a = 1, b = –11, c = 10
b2 – 4ac = (–11)2 – 4(1)(10) Substitution
= 121 – 40 Simplify.
= 81 Subtract.
Answer:
Example 5

45. b2 – 4ac = (–11)2 – 4(1)(10) Substitution = 121 – 40 Simplify.
Describe Roots
B. Find the value of the discriminant for x2 – 11x + 10 = 0. Then describe the number and type of roots for the equation.
a = 1, b = –11, c = 10
b2 – 4ac = (–11)2 – 4(1)(10) Substitution
= 121 – 40 Simplify.
= 81 Subtract.
Answer: The discriminant is 81, so there are two rational roots.
Example 5

46. A. Find the value of the discriminant for x2 + 8x + 16 = 0
A. Find the value of the discriminant for x2 + 8x + 16 = 0. Describe the number and type of roots for the equation.
A. 0; 1 rational root
B. 16; 2 rational roots
C. 32; 2 irrational roots
D. –64; 2 complex roots
Example 5

47. A. Find the value of the discriminant for x2 + 8x + 16 = 0
A. Find the value of the discriminant for x2 + 8x + 16 = 0. Describe the number and type of roots for the equation.
A. 0; 1 rational root
B. 16; 2 rational roots
C. 32; 2 irrational roots
D. –64; 2 complex roots
Example 5

48. B. Find the value of the discriminant for x2 + 2x + 7 = 0
B. Find the value of the discriminant for x2 + 2x + 7 = 0. Describe the number and type of roots for the equation.
A. 0; 1 rational root
B. 36; 2 rational roots
C. 32; 2 irrational roots
D. –24; 2 complex roots
Example 5

49. B. Find the value of the discriminant for x2 + 2x + 7 = 0
B. Find the value of the discriminant for x2 + 2x + 7 = 0. Describe the number and type of roots for the equation.
A. 0; 1 rational root
B. 36; 2 rational roots
C. 32; 2 irrational roots
D. –24; 2 complex roots
Example 5

50. Concept

51. End of the Lesson